Question

solutions to observe that buffers resist pH changes. Burette readings should be made to the nearest 0.1 mL (or 0.05 mL if possible), A. Preparation of Acetic Acid-Acetate Buffer Solution An acetate buffer contains the acid-base pair, acetic acid and the acetate ion (typically added as sodium acetate). For acetic acid, pK, = -log (1.8 x 10-) = 4.74. Consider the con- ditions that will yield 100 mL of an acetate buffer solution with a pH < 4.74. Equation (7) indicates that if the log term is to be zero, the concentrations of the conjugate acid (acetic acid) and the conjugate base (acetate ion) must be equal. Because the conjugate base, sodium acetate, will be dissolved in the same volume that contains the 10 mmol of acetic acid, we must add 10 mmol of the sodium acetate to the acid. The resulting buffer volume may be slightly larger than the original 100 mL of acid because of the volume taken up by the sodium acetate in solution. Students should be aware that sodium acetate may be available as the "anhydrous" salt (CH,COONA) or as a hydrated salt (CH,COONa nH,0). The 10 mmol mass of sodium acetate must be determined using the correct molecular weight. 1. Measure 100 mL of 0.10 M acetic acid in a graduated cylinder. 2. Transfer it to a 250-mL beaker and measure and record the pH. The sodium acetate will be added to this solution. Save it and do not contaminate it. 3. Calculate the quantity of sodium acetate needed to make the acetate buffer. Record your calculations in lab notebook. Make sure to obtain the correct information as to the sodium acetate from the label on the bottle.

With this information what is the PH of the .1 M acetic acid sol and .1 M acetic acid buffer sol? Please help

Answer 1

1) pKa of acetic acid = 4.74

acetic acid is weak acid

Using

pH = 1/2 * [pKa - log C]

or

pH = 1/2 * [4.74 - log (0.1)]

or

pH = 2.87

thus

pH of 0.1 M acetic acid solution = 2.87

2) mmole of acetic acid = 100 ml * 0.10 mole / L = 10.0 mmole.

and

mmole of sodium acetate = 10 mmole.

Using Henderson equation,

pH = pKa + log [salt] / [acid]

or

pH = 4.74 + log (10.0 / 10.0)

or

pH = 4.74

pH of buffer solution = 4.74